Question: The house Trevor's family lives in has $6$ people (including Trevor) and $3$ bathrooms. In the past month, each person showered for an average of $480$ minutes and used an average $72$ liters of shower water (over the entire month). Water costs $0.20$ dollars per liter. How much did Trevor's family pay per minute on shower water?
Solution: There can be many ways to solve this problem. Here, we will do this by thinking about units. Let's say Trevor's family paid for water at a rate of $x\,\dfrac{\text{dollars}}{\text{minute}}$. We are given that the price of water is $0.20\,\dfrac{\text{dollars}}{\text{liter}}$. How can we relate these two quantities with an equation? $\begin{aligned} y\,\dfrac{\text{liters}}{\text{minute}}\cdot 0.2\,\dfrac{\text{dollars}}{\text{liter}}=x\,\dfrac{\text{dollars}}{\text{minute}} \end{aligned}$ So in order to find $x$, the cost per minute, we need to figure out the value of $y$, which is the rate of liters used per minute of shower. Notice what other information we are given: $6\,\text{persons}$ $3\,\text{bathrooms}$ $480\,\dfrac{\text{minutes}}{\text{person}}$ $72\,\dfrac{\text{liters}}{\text{person}}$ Which of these quantities can help us calculate a quantity whose units are $\dfrac{\text{liters}}{\text{minute}}$ ? We can combine the following quantities: $\begin{aligned} &\phantom{=}\dfrac{72\,\dfrac{\text{liters}}{\text{person}}}{480\,\dfrac{\text{minutes}}{\text{person}}} \\\\ &=\dfrac{72}{480}\,\dfrac{\text{liters}}{\cancel\text{person}}\cdot\dfrac{\cancel\text{persons}}{\text{minute}} \\\\ &=0.15\,\dfrac{\text{liters}}{\text{minute}} \end{aligned}$ Now we can plug that in the original equation: $\begin{aligned} 0.15\,\dfrac{\text{liter}}{\text{minute}}\cdot 0.2\,\dfrac{\text{dollars}}{\text{liter}}&=x\,\dfrac{\text{dollars}}{\text{minute}} \\\\ 0.03\,\dfrac{\text{dollars}}{\text{minute}}&=x\,\dfrac{\text{dollars}}{\text{minute}} \end{aligned}$ In conclusion, Trevor's family paid $0.03$ dollars per minute for shower water in the past month.